What is the integral of 3/4 sec(v)tan(v) ?

The integral of 3/4 sec(v)tan(v) is 3/4 sec(v)+C

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integral of 3/4 sec(v)tan(v)

\int \frac{3}{4} sec (v) tan (v) d v

\frac{3}{4} sec (v) + C

Solution steps

\int \frac{3}{4} sec (v) tan (v) d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{3}{4} \cdot \int sec (v) tan (v) d v

Apply u-substitution

: \int 1 d u

= \frac{3}{4} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{3}{4} \cdot 1 \cdot u

Substitute back

u = sec (v)

= \frac{3}{4} \cdot 1 \cdot sec (v)

Simplify

= \frac{3}{4} sec (v)

Add a constant to the solution

= \frac{3}{4} sec (v) + C

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What is the integral of 3/4 sec(v)tan(v) ?
The integral of 3/4 sec(v)tan(v) is 3/4 sec(v)+C